A Fourth Alkene Addition Pattern – Free Radical Addition
Last updated: November 28th, 2022 |
Free Radical Addition Of HBr To Alkenes With ROOR (Peroxides)
We’ve seen that there are three major alkene reactivity patterns [carbocation, three membered ring, and concerted], but there are two minor pathways as well. This post discusses one of them: free-radical addition of HBr to alkenes, which shows the opposite regioselectivity (anti-Markovnikov) than “normal” addition of HBr to alkenes (Markovnikov) which follows the “carbocation” pathway.
Table of Contents
- Free Radical Addition Of HBr To Alkenes Leads To “Anti-Markovnikov” Products
- An Outline Of The Free Radical Mechanism For Addition Of HBr To Alkenes In The Presence Of Peroxides
- Initiation Of The Free-Radical Process Through Homolytic Cleavage Of Peroxides By Heat Or Light
- Formation Of The Bromine Radical From The Alkoxide Radical And HBr
- Propagation Step #1 : Addition Of Bromine Radical To The Alkene Occurs So As To Give The Most Stable Carbon Radical
- Propagation Step #2: The Resulting Carbon Radical Removes A Hydrogen Atom From H–Br, Regenerating The Bromine Radical
- The Termination Step
- Summary: Free-Radical Addition Of HBr To Alkenes
- (Advanced) References and Further Reading
1. Free Radical Addition Of HBr To Alkenes Leads To “Anti-Markovnikov” Products
As discussed previously, alkenes normally react with HBr to give products of “Markovnikov” addition; the bromine ends up on the most substituted carbon of the alkene, and the hydrogen ends up on the least substituted carbon. However, something interesting happens when the same reaction is performed in the presence of peroxides and heat / light: the pattern of addition changes!
Instead of Br ending up on the most substituted carbon of the alkene, it ends up on the least. [The stereochemistry of the reaction, however, is unchanged: it still gives a mixture of “syn” and “anti” products.]
This so-called “anti-Markovnikov” addition is intriguing. What difference could the presence of peroxides, and furthermore heat (or light) make to this reaction?
2. An Outline Of The Free Radical Mechanism For Addition Of HBr To Alkenes In The Presence Of ROOR (Peroxides)
This reaction occurs through a free-radical process. (For a primer on free radical chemistry, you might want to check out this introductory article on Free Radical Reactions). Here is an outline of the mechanism:
- Peroxides contain a weak oxygen-oxygen bond [approximately 35 kcal/mol; compare to C-H at approx 100 kcal/mol]
- Heating leads to homolytic fragmentation of this bond – that is, the bond breaks such as to leave one unpaired electron on each atom. Strong sources of light [e.g. a floodlight or other source of light radiation which reaches into the near UV] can also serve to sever this bond.
- The resulting highly reactive alkoxy radical can then abstract a hydrogen from H-Br, giving a bromine radical. The bromine radical is the species that adds to the alkene.
- Addition to the alkene will preferably occur in such a way that the most stable free radical is formed [in the case above, the tertiary radical]. That’s why bromine ends up on the least substituted carbon of the alkene. (See: 3 Factors Which Stabilize Free Radicals)
- This tertiary radical then removes hydrogen from H-Br, liberating a bromine radical, and the cycle continues.
3. Initiation Of The Free-Radical Process Through Homolytic Cleavage Of ROOR (Peroxides) By Heat Or Light
Only a trace [catalytic] amount of peroxide is required to get the reaction started, although of course at least one molar equivalent of HBr is required to result in full addition of HBr to the alkene.
In the first step, addition of energy (in the form of heat or light) leads to homolytic fragmentation of the weak O–O bond to generate two new free radicals. “Homolytic” means that the bond is broken such that each atom receives the same (“homos” = Greek for “same”) number of electrons.
(Most of the bond breakage we see in organic chemistry is heterolytic, where the bond breaks unequally. )
The singly barbed arrows depict the movement of single electrons; two alkoxy radicals are formed. Since there is a net increase in the number of radicals (0 →2) this is an initiation step.
Common “peroxides” for this purpose are t-butyl peroxide or benzoyl peroxide. [Note 1]. Alternatively other free-radical “initiators” such as AIBN can also be used.
Only a catalytic amount of peroxides are used to initate this reaction (typically 10-20 mole %, although more can be used, especially when added batchwise).
4. Formation Of The Bromine Radical From The Alkoxide Radical And HBr
In the next step, one of the oxygen radicals from step 1 removes a hydrogen from H–Br in another homolytic process.
Here, we’re forming an H–O bond (bond dissociation energy of 102 kcal/mol for H–O in CH3OH) and breaking an H–Br bond (bond dissociation energy of 87 kcal/mol) , so a difference in energy of about 15 kcal/mol makes this process essentially irreversible.
(Note: since this process does not change the number of free radicals, it is technically a propagation step)
5. Propagation Step #1 : Addition Of Bromine Radical To The Alkene Occurs So As To Give The Most Stable Carbon Radical
Once formed, the bromine radical can then add to the alkene.
In a relatively “flat” alkene such as 1-methylcyclohexene, addition of the radical will occur with equal probability from either face.
The question is, which atom of the double bond does the free radical attack? The bond could break two different ways, after all.
- Attack of the bromine radical on the more substituted carbon would result in a new free radical on a secondary carbon.
- Attack of the bromine radical on the less substituted carbon would result in a new free radical on a tertiary carbon.
Free radicals are electron-deficient species and are stabilized by adjacent electron donors. The more stable free radical intermediate is the tertiary free radical, and that is why addition occurs predominantly at the less substituted carbon (i.e. the carbon attached to the fewest number of carbons).
This explains the “anti-Markovnikov” selectivity of the reaction.
6. Propagation Step #2: The Resulting Carbon Radical Removes A Hydrogen Atom From H–Br, Regenerating The Bromine Radical
In a second propagation step in the main sequence, the resulting carbon radical removes a hydrogen from another equivalent of H–Br, giving the final addition product.
Alkyl free radicals are sp2-hybridized, and are shallow pyramids that invert easily.
H–Br, therefore, can react on either face of the free radical [Note 2]. If it attacks on the same face as the Br, then we obtain a “syn” product. If it attacks on the opposite face of the Br, then the product is “anti“.
A mixture of both will be obtained. The reaction is not stereoselective.
A bromine radical is generated by this process, which can then add to another equivalent of alkene (propagation step #1).
7. The Termination Step
When the concentration of HBr and alkene become low relative to the concentration of free radical, termination can occur (See post: Initiation, Propagation, Termination). This could occur through a variety of specific pathways (not shown) involving recombination of two free radicals to generate a new bond.
8. Summary: Free-Radical Addition Of HBr To Alkenes
This reaction pathway is most commonly observed (in Org 1 and Org 2, anyway) for addition of HBr, although a rich chemistry of radical addition reactions to alkenes exists (particularly for organostannanes).
NEXT POST: Ozonolysis of Alkenes
Note 1. Benzoyl peroxide enjoys a common household use as an acne cleanser, and even makes an appearance in this classic ad for Oxy skin care (“Oxycute ‘Em”.)
Note 2. The geometry of free radical carbons is that of a shallow pyramid with a low barrier for inversion, allowing for reactivity on either face. The exception is in weird cases where inversion would be highly disfavored, such as on a bridgehead (See Bicyclic Molecules and How To Name Them).
(Advanced) References and Further Reading
Bond Dissociation Energies From Lowry & Richardson, “Mechanism and Theory In Organic Chemistry“, Harper & Row, 1987 pp 161-162
- The Peroxide Effect in the Addition of Reagents to Unsaturated Compounds and in Rearrangement Reactions.
Frank R. Mayo and Cheves Walling
Chemical Reviews 1940, 27 (2), 351-412
F. R. Mayo was a student of the prominent chemist M. S. Kharasch, and together they first described the “peroxide effect” in the anti-Markovnikov addition of HBr to alkenes, ascribing it to a free-radical mechanism.
- ADDITION OF HYDROGEN BROMIDE TO 4,4-DIMETHYLPENTENE-1
M. S. Kharasch, Chester Hannum, and M. Gladstone
Journal of the American Chemical Society 1934, 56 (1), 244-244
This communication describes both products that are obtained from HBr addition to the title olefin via the electrophilic and radical mechanisms.
- THE PHOTO-ADDITION OF HYDROGEN BROMIDE TO OLEFINIC BONDS
WILLIAM E. VAUGHAN, FREDERICK F. RUST, and THEODORE W. EVANS
The Journal of Organic Chemistry 1942, 07 (6), 477-490
The radical addition of HBr can be initiated not just by peroxides, but also by light, as this paper describes.
- The Peroxide Effect in the Addition of Reagents to Unsaturated Substances. XXII. The Addition of Hydrogen Bromide to Trimethylethylene, Styrene, Crotonic Acid, and Ethyl Crotonate
Cheves Walling, M. S. Kharasch, and F. R. Mayo
Journal of the American Chemical Society 1939, 61 (10), 2693-2696
Kharasch and co-workers reported the hydrobromination of styrene in dilute pentane solution with dibenzoyl peroxide to give an 80 : 20 ratio in favor of the primary bromide. Unfortunately, detailed conditions were not provided.
- Scalable anti-Markovnikov hydrobromination of aliphatic and aromatic olefins
Marzia Galli, Catherine J. Fletcher, Marc del Pozo, and Stephen M. Goldup
Org. Biomol. Chem., 2016, 14, 5622-5626
This is an interesting paper demonstrating the relevance of this chemistry in modern organic synthesis; it describes the rediscovery of simple scalable conditions for synthesis of primary bromides under “initiator free” conditions from alkyl and aryl alkenes.
29 thoughts on “A Fourth Alkene Addition Pattern – Free Radical Addition”
Penultimate paragraph: “This reaction pathway is most commonly observed (in Org 1 and Org 2, anyway) for” – unfinished sentence.
Great job, as usual.
Fixed. Thank you for reminding me to put in a brief shout out to organostannane chemistry.
is there a video tutorial on this? I learn better by hearing and doing rather than reading :)
Not at present, sorry!
Why hydrogen free radical does not add first to alkene instead of bromine free radical
Because it is not formed! See the comment below for a clarification of (or at least, an attempt to clarify) it.
Why does the tertiary radical have a preference to remove the hydrogen from the HBr instead of the bromine?
Not sure what you mean – you mean react with bromine radical?
The concentration of bromine radical at any given moment is much less than the concentration of HBr.
If it reacted with bromine radical that would be a termination step!
No, in fact he is asking about the ‘Abstraction of hydrogen in H-Br…’ step, why don’t we get a dibromo cyclo-product. Well, that reaction would simply be thermodynamically unfavorable (+39 kcal in contrast with -23 kcal). This is discussed on page 387 in the book ‘Basic principles of organic chemistry’ (just Google it, it’s fascinating!). And I hope this helps.
In step 2, why does the bromine radical not abstract an allylic hydrogen (like in NBS halogenation reactions)? I presume that allylic radical would be almost as stable as a tertiary radical.
I think it’s because the C-C pi bond strength is about 60-65 kcal/mol whereas C-H allylic bond strength is about 80-85 kcal/mol. Higher activation energy in other words
Could it be that allylic hydrogen abstraction does occur, but the resulting allyl radical has nothing to react with except H-Br? The allyl radical could reabstract a hydrogen from H-Br, to reform the alkene, which then eventually adds a bromine radical. Potentially the alkene could isomerize if this occurred, and I don’t know if this is observed. Just a thought…
Thanks Bruce, I think that’s exactly what’s going on.
Do you have a post for the radical addition of HBr when there are no peroxides, niher words, the other pathway?
The radical pathway requires peroxides. The “ionic” pathway involves carbocations and is covered here: https://www.masterorganicchemistry.com/2013/02/22/addition-pattern-1-the-carbocation-pathway/
Could you comment on the stability of the stereoisomers which result from this reaction? Thank you!
The stability of the stereoisomers? It will be mixture of configurations. Hard to generalize.
How can you determine if a radical reaction is initiated by heat or light?
Either can work. For an example, see here (benzylic bromination of xylene) http://www.orgsyn.org/demo.aspx?prep=CV4P0984
What happens when Zn dust is added to HBr solution before alkene is added?
Do you have a specific example?
is it necessary to learn bond energy vaule in undergraduate studies….
I think so, yes. Is it necessary to learn the value of the hands in poker?
My textbook says that anti-Markovnikov addition only occurs with HBr, and not HCl. Can you please tell me why?
HCL will not form chlorine radical as chlorine is highly electronegative…it will always form a cation only.
This is incorrect.
In the last step, why the tertiary radical react with H from H-Br not Br from H-Br? (and the product will be 1,2-dibromo-2-methylcyclohexane)
Reaction with Br will give the hydrogen radical, H• , which is considerably less stable than the bromine radical. And formation of the C-Br bond is much weaker than formation of the C-H bond.